JEE Main 2021 — Differential Equations Question with Solution
From: JEE Main 2021 (Online) 27th August Morning Shift
Question
Let us consider a curve, y = f(x) passing through the point (2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
(given)
If
Solution of DE
Passes through (2, 2), so
12 = 8 + c c = 4
3xy = x3 4
i.e. 3x . f(x) = x3 4
If
Solution of DE
Passes through (2, 2), so
12 = 8 + c c = 4
3xy = x3 4
i.e. 3x . f(x) = x3 4
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This is a previous-year question from JEE Main 2021, covering the Differential Equations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.